In a Young's double slit experiment, the intensities at two points, for the path differences λ/4 and λ/3 ( λ being the wavelength of light used) are I₁ and I₂ respectively. If I₀ denotes the intensity produced by each one of the individual slits, then (I₁+I₂)/I₀= __________.

Solution

Physics sir · 4 steps

Turn each path difference into a phase difference, then use I = 4I₀cos²(φ/2) for each point.

1. Intensity formula. Two slits of intensity I₀ each give a total intensity that depends on the phase difference φ between the waves. At a bright fringe it reaches 4I₀.

   I = 4I₀cos²(φ/2), φ = 2π/λ Δ x

2. First point. For Δ x = λ/4, φ = π/2, so φ/2 = π/4 and cos² π/4 = 1/2. This gives I₁ = 4I₀ × 1/2 = 2I₀.
3. Second point. For Δ x = λ/3, φ = 2π/3, so φ/2 = π/3 and cos² π/3 = 1/4. This gives I₂ = 4I₀ × 1/4 = I₀.
4. So. I₁ + I₂ = 2I₀ + I₀ = 3I₀, so (I₁ + I₂)/I₀ = 3.

The common slip: 0.75. Writing I = I₀cos²(φ/2) treats I₀ as the peak intensity and drops the factor 4, which gives 1/2 + 1/4 = 0.75.

The idea to keep. In YDSE, the maximum intensity is 4I₀, not I₀. Convert path difference to phase with φ = 2π/λΔ x, and remember the cosine takes half of that phase.

Answer from the NTA final answer key. Solution by PrepWiser.